[seqfan] Draft proof

[Jonathan Stauduhar]

Hi folks,

Late last year I corrected a typo in <https://oeis.org/A007921>A 
<https://oeis.org/A005381>007921.   I changed "Equals A005381 
<https://oeis.org/A005381>(2n-1) - 1" to "Equals A068780 
<https://oeis.org/A068780>(2n-1) - 1".  Neil asked if I had a proof that 
this was true, but I did not.  Here is a first attempt at that proof.  
I'm no proof writer, so doubtless there are errors and style flaws -- 
feedback appreciated.


We want to show that A007921(n) = A068780(2n-1) - 1.

For this to be true we need to prove the following:

(1) That every term in A007921 is odd.
(2) A007921(n) + 1 and A007921(n) + 2 are both composite numbers.


Proof of (1):

   Let p and q be prime, where q < p. There are two cases:  q = 2; q > 
2.  With q = 2, p - q is odd for all p.  For q > 2, since p and q are 
odd numbers, the result of p - q is always an even number.  Since no 
even number can appear in A007921 every term is odd.


Proof of (2):

Lemma 1:  Since 2 is prime, no lesser of a twin prime pair can appear in 
A007921.

(3) A007921 + 1 is composite:

Proof of (3):

   Since (1) is true, clearly, adding 1 to any term in A007921 produces 
an even (composite) number.


(4) A007921 + 2 is composite:

Proof of (4):

   There are two cases:  (a) A007921(n) is prime;  (b) A007921(n) is 
composite odd.

For case (a):  By (Lemma 1), adding 2 to any prime term must result in a 
composite odd.

For case (b):  there are 2 cases:  (c) adding 2 to a composite odd 
results in another composite odd; (d) adding 2 to a composite odd 
results in a prime.

(c) is self-evident.

Proof of (d):

   Let c be a composite odd and let q = 2.  Then c + q = p. Rewrite and 
we have p - c = q = 2, which implies c is the lesser of a twin prime 
pair, which is a contradiction because c is composite.  So case (b) can 
never happen.

Q.E.D.